Problem: If $x \bigtriangledown y = 2x^{2}-y^{2}$ and $x \odot y = 4x-y$, find $0 \odot (-4 \bigtriangledown 4)$.
Solution: First, find $-4 \bigtriangledown 4$ $ -4 \bigtriangledown 4 = 2(-4)^{2}-4^{2}$ $ \hphantom{-4 \bigtriangledown 4} = 16$ Now, find $0 \odot 16$ $ 0 \odot 16 = (4)(0)-16$ $ \hphantom{0 \odot 16} = -16$.